Brain Gains (Rapid Recall, Jivin' Generation)

  • Consider the function $z=\sin(x)+e^y$, where $x=3t$ and $y=t^2$. Compute $\frac{dz}{dt}$.

Solution

There are two ways to do this.

  1. Substitution gives $z=\sin(3t)+e^{t^2}$. Differentiation (using the chain rule) then gives $$\frac{dz}{dt}=\cos(3t)\frac{d}{dt}(3t)+e^{t^2}\,\frac{d}{dt}(t^2)=\cos(3t)3+e^{t^2}\,2t.$$
  2. Differentials give $dz = \cos(x)dx+e^ydy$, with $dx = 3dt$ and $dy=2tdt$. Substitution then gives $$dz = \cos(3t)3dt+e^{t^2}\,2tdt\quad\text{or}\quad\frac{dz}{dt}=\cos(3t)3+e^{t^2}\,2t.$$

In both cases, we obtained the same solution of $$\frac{dz}{dt}=\underbrace{\cos(3t)}_{f_x}\underbrace{3}_{\frac{dx}{dt}}+\underbrace{e^{t^2}}_{f_y}\underbrace{2t}_{\frac{dy}{dt}}.$$ Writing the solution above symbolically gives us the chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$

  • Suppose $dz = e^{x^2}dx+\cos(2y)dy$, $x=3t$, and $y=t^2$. Compute $\frac{dz}{dt}$.

Solution

This time we don't know what the function $z$ equals, so we cannot first substitute and then differentiate. We do know however that $f_x = e^{x^2}$ and $f_y = \cos(2y)$. We can compute differentials and then substitute.

  • Note that $dx = 3dt$ and $dy = 2tdt$. Substitution then gives $$dz = e^{(3t)^2}3\,dt+\cos(2(t^2))2t\,dt \quad\text{and so}\quad \frac{dz}{dt} = e^{(3t)^2}3+\cos(2(t^2))2t.$$

Again, the the solution above symbolically gives us the same chain rule $$\frac{dz}{dt} = f_x\frac{dx}{dt}+f_y\frac{dy}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$$

  • For the function $f(x,y) = x^2y^4$, we have $\vec \nabla f(x,y) = (2xy^4,4x^2y^3)$. Compute the differential of $\vec \nabla f(x,y)$, and then state the second derivative $D^2f(x,y)$.

Solution

We have $$\begin{align} d(\vec\nabla f) &= d(2xy^4,4x^2y^3)\\ &= ((2dx)y^4+2x(4y^3dy),(8xdx)y^3+4x^2(3y^2dy))\\ &= \begin{pmatrix}(2)y^4\\(8x)y^3\end{pmatrix}dx+\begin{pmatrix}2x(4y^3)\\4x^2(3y^2)\end{pmatrix}dy\\ &= \begin{bmatrix}(2)y^4&2x(4y^3)\\(8x)y^3&4x^2(3y^2)\end{bmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}. \end{align}$$ The last two lines above give the differential of $\vec \nabla f$ as a linear combination of partial derivatives, and then as a matrix product. The last line show the second derivative of $f$ is $$D^2f(x,y) = \begin{bmatrix}(2)y^4&2x(4y^3)\\(8x)y^3&4x^2(3y^2)\end{bmatrix} = \begin{bmatrix}2y^4&8xy^3\\8xy^3&12x^2y^2\end{bmatrix} .$$ We can also obtain this matrix by just directly computing all the second partial derivatives.

  • $f_x = 2xy^4$ which means $f_{xx} = 2y^4$ and $f_{xy} = 8xy^3$. These form the first column.
  • $f_y = 4x^2y^3$ which means $f_{yx} = 8xy^3$ and $f_{yy} = 12x^2y^2$. These form the second column.

Group Problems

  1. A rover travels along the line $g(x,y)=2x+3y=6$. The surrounding terrain has elevation $f(x,y)=x^2+4y$. The rover reaches a local minimum along this path, and our job is to find the location of this minimum.
    • Compute $\vec \nabla f$ and $\vec \nabla g$.
    • Write the system of equations that results from $\vec \nabla f=\lambda\vec \nabla g$ together with $g(x,y) = 6$.
    • Solve the system above (you should get $x=4/3$ and $y=10/9$).
    • Use LagrangeMultipliers.nb to check your work and visualize the rover's path and how it relates to the elevation contours. Scroll down to the "All Code in One Block" section, and update f, g, and c.
  2. For the function $f(x,y)=x^2+4xy+3y^2-10x-18y$, verify that the first derivative $Df(x,y)$ and second derivative $D^2f(x,y)$ are $$Df(x,y) = \begin{bmatrix}2x+4y-10&4x+6y-18\end{bmatrix}\quad\text{and}\quad D^2f(x,y) = \begin{bmatrix}\begin{matrix}2\\4\end{matrix}&\begin{matrix}4\\6\end{matrix}\end{bmatrix}. $$
    • Solve $Df(x,y)=\begin{bmatrix}0&0\end{bmatrix}$, to find the critical points of this function. [Check: $(x,y)=(3,1)$.]
    • Find the eigenvalues of $D^2f(3,1)$. [Check: $\lambda = 4\pm\sqrt{20} = 4\pm 2\sqrt{5}$.]
    • Does the function $f$ have a local max, local min, or saddle at $(3,1)$? Explain.
  3. Consider the function $f(x,y,z) = 3xy+z^2$. We'll be analyzing the level surface that passes through the point $P=(1,-3,2)$.
    • Compute the differential $df$, and then evaluate the differential at $P$.
    • For a level surface, the output remains constant (so $df=0$). If we let $(x,y,z)$ be a point on the surface really close to $P$, then we have $dx=x-1$, $dy=y-(-3)$ and $dz = z-?$. Plug this information into the differential at $P$ to obtain an equation of the tangent plane.
    • Give an equation of the tangent plane to the level surface of $f$ that passes through $(1,2,-3)$.
    • Give an equation of the tangent plane to the level surface of $f$ that passes through $(a,b,c)$.
    • What relationship exists between the gradient of $f$ at $P$ and the tangent plane through $P$?
  4. Suppose a plane passes through the point $(a,b,c)$ and has normal vector $(A,B,C)$. Give an equation of that plane.
  5. Give an equation of the tangent plane to $xy+z^2=7$ at the point $P=(-3,-2,1)$.
  6. Give an equation of the tangent plane to $z=f(x,y)=xy^2$ at the point $P=(4,-1,f(4,-1))$.